$(\csc^2\theta-1)(\tan^2\theta) = \; ?$
Solution: We can derive a useful identity from ${\sin^2 \theta} + {\cos^2 \theta} = 1$ to simplify this expression. $1$ ${\sin\theta}$ ${\cos\theta}$ $\theta$ We can see why this identity is true by using the Pythagorean Theorem. Dividing both sides by $\sin^2\theta$ , we get $ \dfrac{\sin^2\theta}{\sin^2\theta} + \dfrac{\cos^2\theta}{\sin^2\theta} = \dfrac{1}{\sin^2\theta}$ $ 1 + \cot^2\theta = \csc^2\theta$ $ \csc^2\theta-1 = \cot^2\theta$ Plugging into our expression, we get $ (\csc^2\theta-1)(\tan^2\theta) = \left(\cot^2\theta\right) \left(\tan^2\theta\right) $ To make simplifying easier, let's put everything in terms of $\sin$ and $\cos$ . We know $\cot^2\theta = \dfrac{\cos^2\theta}{\sin^2\theta}$ and $\tan^2\theta = \dfrac{\sin^2\theta}{\cos^2\theta}$ , so we can substitute to get $ \left(\cot^2\theta\right) \left(\tan^2\theta\right) = \left(\dfrac{\cos^2\theta}{\sin^2\theta}\right) \left(\dfrac{\sin^2\theta}{\cos^2\theta}\right) $ This is $1$.